What is the integral from-2 to 0 of 9x^2(2x^3+3)^2 ?

The integral from-2 to 0 of 9x^2(2x^3+3)^2 is 1112

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integral from-2 to 0 of 9x^2(2x^3+3)^2

\int_{- 2}^{0} 9 x^{2} (2 x^{3} + 3)^{2} d x

1112

Solution steps

\int_{- 2}^{0} 9 x^{2} (2 x^{3} + 3)^{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 9 \cdot \int_{- 2}^{0} x^{2} (2 x^{3} + 3)^{2} d x

Apply u-substitution

: \int_{- 8}^{0} \frac{( 2 u + 3 ) ^{2}}{3} d u

= 9 \cdot \int_{- 8}^{0} \frac{( 2 u + 3 ) ^{2}}{3} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 9 \cdot \frac{1}{3} \cdot \int_{- 8}^{0} (2 u + 3)^{2} d u

Expand

(2 u + 3)^{2} : 4 u^{2} + 12 u + 9

= 9 \cdot \frac{1}{3} \cdot \int_{- 8}^{0} 4 u^{2} + 12 u + 9 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 9 \cdot \frac{1}{3} (\int_{- 8}^{0} 4 u^{2} d u + \int_{- 8}^{0} 12 u d u + \int_{- 8}^{0} 9 d u)

\int_{- 8}^{0} 4 u^{2} d u = \frac{2048}{3}

\int_{- 8}^{0} 12 u d u = - 384

\int_{- 8}^{0} 9 d u = 72

= 9 \cdot \frac{1}{3} (\frac{2048}{3} - 384 + 72)

Simplify

9 \cdot \frac{1}{3} (\frac{2048}{3} - 384 + 72) : 1112

= 1112

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What is the integral from-2 to 0 of 9x^2(2x^3+3)^2 ?
The integral from-2 to 0 of 9x^2(2x^3+3)^2 is 1112